导体和电介质静电学
$$\frac{\partial \small{物理量}}{\partial t}=0\qquad\boldsymbol{j}=0\\
W=\frac{1}{2}\int \boldsymbol{D}\cdot\boldsymbol{E}\dd V=-\frac{1}{2}\int \boldsymbol{D}\cdot\nabla\varphi\dd V\\=-\frac{1}{2}\int [\nabla\cdot(\varphi\boldsymbol{D})-\varphi\nabla\cdot\boldsymbol{D}]\dd V\\
=-\frac{1}{2}\oint \varphi\boldsymbol{D}\cdot\dd\boldsymbol{S}+\frac{1}{2}\int \varphi\rho\dd V\\
W_t=\frac{1}{2}\int_\infty \boldsymbol{D}\cdot\boldsymbol{E}\dd V=\frac{1}{2}\int \varphi\rho\dd V$$
$$
\boldsymbol{E}_内=\frac{\boldsymbol{j}}{\sigma_c}=0\qquad\rho_内=\nabla\cdot\boldsymbol{D}_内=\varepsilon\nabla\cdot\boldsymbol{E}_内=0\\
\hat{\boldsymbol{n}}\times\boldsymbol{E}_{外表面}=0\qquad\hat{\boldsymbol{n}}\cdot\boldsymbol{D}_{外表面}=\sigma\\
\varphi|_{边界}=\small{常数}\qquad\normalsize \frac{\partial\varphi}{\partial n}|_{边界}=-\frac{\sigma}{\varepsilon}\\
\delta W_t\overset{电荷分布变动}{=}-\varepsilon\int_\infty\nabla\varphi\cdot\delta\boldsymbol{E}\dd V\\
=-\varepsilon\int_\infty[\nabla\cdot(\varphi\delta\boldsymbol{E})-\varphi\nabla\cdot\delta\boldsymbol{E}]\dd V\\
=-\varepsilon\oint_\infty\varphi\delta\boldsymbol{E}\cdot\dd\boldsymbol{S}+\varepsilon\int\varphi\delta\rho\dd V\\
=\varepsilon\sum_i\varphi_i\delta Q_i=0$$
$$W_t=\frac{1}{2}\int_\infty\varphi\rho\dd V=\frac{1}{2}\sum_i\varphi_iQ_i\\
\varphi_i=C_{ij}^{-1}Q_j\qquad Q_i=C_{ij}\varphi_j\\
C_{ij}=C_{ji}\qquad C_{ij}^{-1}=C_{ji}^{-1}\\
W=\frac{1}{2}\sum_{ij}C_{ij}\varphi_i\varphi_j=\frac{1}{2}\sum_{ij}C_{ij}^{-1}Q_iQ_j>0\\
\det\mathbb{C}>0\qquad\det\mathbb{C}^{-1}>0\\
C_{ii}>0\qquad C_{ij}<0\;(i\neq j)\\
W_{两体}=\frac{\varepsilon}{2}\int E^2\dd V\\=\frac{\varepsilon}{2}\int E_1^2\dd V+\frac{\varepsilon}{2}\int E_2^2\dd V+\varepsilon\int \boldsymbol{E}_1\cdot\boldsymbol{E}_2\dd V\\
W_1=\frac{\varepsilon}{2}\int E_1^2\dd V\qquad W_2=\frac{\varepsilon}{2}\int E_2^2\dd V\qquad W_{12}=\varepsilon\int \nabla\varphi_1\cdot\nabla\varphi_2\dd V$$
$$W_i=\frac{1}{2}\sum_{\alpha}q_\alpha\varphi_\alpha\\
\nabla^2 W_i=0\Rightarrow\small 不能稳定平衡$$
$$\int_{\infty-V_{导体}}[\nabla\cdot(\varphi\nabla\varphi’)-\nabla\cdot(\varphi\nabla\varphi’)]\dd V=\int_{\infty-V_{导体}}(\varphi\nabla^2\varphi’-\varphi’\nabla^2\varphi)\dd V\\
=0=\oint_S(\varphi\nabla\varphi’-\varphi’\nabla\varphi)\cdot\dd\boldsymbol{S}\\
=\oint_\infty(\varphi\nabla\varphi’-\varphi’\nabla\varphi)\cdot\dd\boldsymbol{S}-\sum_i\oint_{S_i}(\varphi_i\nabla\varphi’_i-\varphi’_i\nabla\varphi_i)\cdot\dd\boldsymbol{S}_i\\
=-\sum_i\oint_{S_i}(\varphi_i\frac{\partial\varphi_i’}{\partial n}-\varphi’_i\frac{\partial\varphi_i}{\partial n})\dd S_i=\frac{1}{\varepsilon}\sum_i(\varphi_i Q_i’-\varphi_i’ Q_i)\\
\sum_i\varphi_i Q_i’=\sum_i\varphi_i’ Q_i$$
$$\dd\boldsymbol{F}=-\mathbb{T}\cdot\dd\boldsymbol{S}\\
\frac{\dd\boldsymbol{F}}{\dd S}=-\hat{\boldsymbol{n}}\cdot\mathbb{T}=\frac{\varepsilon}{2}E^2\hat{\boldsymbol{n}}=\frac{1}{2\varepsilon}\sigma^2\hat{\boldsymbol{n}}$$
$$\delta W_t\overset{介质位移\boldsymbol{\xi}}{=}\frac{1}{2}\delta\int_\infty\frac{D^2}{\varepsilon}\dd V\\
=\frac{1}{2}\int_\infty(2\boldsymbol{E}\cdot\delta\boldsymbol{D}-E^2\delta\varepsilon)\dd V\\
=\frac{1}{2}\int_\infty(-2\nabla\cdot(\varphi\delta\boldsymbol{D})+2\varphi(\nabla\cdot\delta\boldsymbol{D})-E^2\delta\varepsilon)\dd V\\=\frac{1}{2}\int_\infty(2\varphi\delta\rho-E^2\delta\varepsilon)\dd V\\
\int_V\delta\rho\dd V=-\oint_S\rho\boldsymbol{\xi}\cdot\dd\boldsymbol{S}=-\int_V\nabla\cdot(\rho\xi)\dd V\\
\delta\rho=-\nabla\cdot(\rho\boldsymbol{\xi})\\
\delta W_t=\frac{1}{2}\int_\infty[-2\varphi\nabla\cdot(\rho\boldsymbol{\xi})+E^2\frac{\dd\varepsilon}{\dd\rho_{介质}}\nabla\cdot(\rho_{介质}\boldsymbol{\xi})]\dd V\\
=\frac{1}{2}\int_\infty[2\nabla\varphi\cdot(\rho\boldsymbol{\xi})-\nabla(E^2\frac{\dd\varepsilon}{\dd\rho_{介质}})\cdot(\rho_{介质}\boldsymbol{\xi})]\dd V\\
\boldsymbol{f}=-\frac{\delta w}{\boldsymbol{\xi}}=\rho\boldsymbol{E}+\frac{\rho_{介质}}{2}\nabla(E^2\frac{\dd\varepsilon}{\dd\rho_{介质}})\\
=\rho\boldsymbol{E}-\frac{1}{2}E^2\frac{\dd\varepsilon}{\dd\rho_{介质}}\nabla\rho_{介质}+\frac{1}{2}\nabla(E^2\frac{\dd\varepsilon}{\dd\rho_{介质}}\rho_{介质})\\
=\rho\boldsymbol{E}-\frac{1}{2}E^2\nabla\varepsilon+\frac{1}{2}\nabla(E^2\frac{\dd\varepsilon}{\dd\rho_{介质}}\rho_{介质})$$